series problems for STD V and STD VI

Questions on the application of the formulae:

For S = 1 + 2 + 3 + .....n terms = i.e. sum of natural numbers from 1 to n.
S = n (n+1)/2

1. S = ( 1 + 3 + 5 + ... 25 terms) + ( 2 + 4 + 6 ..... 25 terms)
2. S = (3 + 6 + 9 ...... 25 terms)
3. S = ( 2 + 4 + 6 + ...... 25 terms)
4. S = (1 - 1/5) + (1 - 2/5) + (1 - 3/5) + (1 - 4/5) + ..... 25 terms
5. S = (1 - 1/n) + (1 - 2/n) + (1 - 3/n) +........ n terms
6. S = (2a - 3b) + (4a - 6b) + (6a - 9b) +.......... 25 terms, given a = 5 and b = 2

Answers:

1. S = 1+ 2 + 3 ... 25 terms ( after re-arranging the terms). So, S = 25 x 26/2 = 325
2. S = 3( 1 + 2 + 3 ... 25 terms) So, S = 3 x 25 x 26/2 = 975
3. S = 2( 1+ 2 + 3... 25 terms) So, S = 2 x 25 x 26/2 = 650
4. S = ( 1 + 1 ... 25 terms) - 1/5(1 + 2 + ... 25 term) = 25 - (1/5) x 25 x 26/2 = - 40
5. S = ( 1 + 1 ... n terms) - 1/n(1 + 2 + ... n term) = n - (1/n) x n x (n +1)/2 = (n - 1)/2
6. S = 2a(1 + 2 +... 25 terms) - 3b(1 + 2 + ... 25 terms) = 2a x 325 - 3b x 325 = 1300